Now clearly the odd terms in this sequence are negative while the even terms of this sequence are positive. But it's not clear to me why this sequence will always be bounded by $1$. For example, if we took out the negative terms the sequence could certainly be larger than $1$ (so we can't bound $e^x$ by its positive terms).

What then are then some useful bounds with the exponential function to solve problems like this and related problems? What if we fixed $t > 0$ and considered for example $e^{tx}$?

Also, you just need to note that the function is monotonically increasing because each term of the series expansion is, and then note that $e^0=1$. This gives the upper bound.
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PotatoOct 29 '12 at 7:07

The power series forthe exponential function should also have a constant term.
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Per ManneOct 29 '12 at 7:09

By the way, all of this is done rigorously in the relevant chapter of Ahlfors. I'm sure there's a more accessible source, but I can't think of one off the top of my head.
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PotatoOct 29 '12 at 7:14

1 Answer
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Short answer: If you believe (*) $e^{x+y} = e^x e^y$, in particular for any $a > 0$ we have $e^{a + -a} = 1 \Rightarrow e^{-a} = \frac{1}{e^a}$, so it suffices to show $e^{a} > 1$ which is clear (first term is already 1, rest are all positive in the power series).

Slightly more: a big motivation for introducing the exponential function at all is that a priori real numbers under addition, and positive reals under multiplication are very different things. $exp$ furnishes an "isomorphism" of "groups" between them that's continuous and normalized nicely. So it helps (me at least) to keep in mind the "purpose" of the exponential function, even when working with it rigorously.

To see ($*$), there's a lovely fact that power series converge absolutely where they converge at all, so you're free to shuffle terms around in the most naive way. With this, I claim you can check (*) directly by binomially expanding $(x+y)^k$ in each term of $e^{x+y}$. It's a good exercise! Hope that helps.